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Here's the problem.

$$\lim_{x\to 0} \frac{\sin x(1 - \cos x)}{x^2}$$

I really don't know where to start with this. Please help.

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    $\begingroup$ Do you know L'Hopital's rule, where you can differentiate the numerator, and the denominator, so long as you have an indeterminate form? Here, you have an indeterminate form as is: $\frac 00$. $\endgroup$ – Namaste Sep 25 '13 at 14:26
  • $\begingroup$ Use Maple to this end. $\endgroup$ – user64494 Sep 25 '13 at 14:37
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    $\begingroup$ Well we haven't learned about derivatives in class yet, but after watching a video, L'Hopital's Rule seems very useful. Thank you $\endgroup$ – Steve Sep 25 '13 at 14:40
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I assume you know that $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$ Now, $$\lim_{x \to 0} \frac{1-\cos x}{x^2} = \lim_{x \to 0} \frac{(1-\cos x)(1+\cos x)}{x^2 (1+\cos x)}=\lim_{x \to 0} \frac{\sin^2 x}{x^2(1+\cos x)},$$ and therefore $$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 \times\lim_{x \to 0}\frac{1}{1+\cos x}=\frac{1}{2}.$$ Finally, $$\lim_{x \to 0} \frac{\sin x (1-\cos x)}{x^2}=\lim_{x \to 0} \sin x \times \lim_{x \to 0} \frac{1-\cos x}{x^2}=0 \times \frac{1}{2}=0.$$

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Hint: $$\lim_{x\to 0}\frac{\sin x(1-\cos x)}{x^2} = \left(\lim_{x\to 0}\frac{\sin x - \sin 0}{x-0}\right)(-1)\left(\lim_{y\to 0}\frac{\cos y - \cos 0}{y-0}\right),$$ assuming the two limits on the right hand side exist (why do they exist?).

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  • $\begingroup$ Hmmm...sorry I'm kinda lost on how you came up with the right side. $\endgroup$ – Steve Sep 25 '13 at 14:45
  • $\begingroup$ @Steve: $\frac{\sin x(1-\cos x)}{x^2} = \frac{\sin x}{x}\cdot\frac{1-\cos x}{x} = \frac{\sin x-\sin 0}{x-0}\cdot(-1)\cdot\frac{\cos x-\cos 0}{x-0}$ $\endgroup$ – Jonathan Y. Sep 25 '13 at 14:54
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Applying the l'Hopital rule, if you put: $f=\frac{\sin(x)(1-\cos(x))}{x^2}=\frac{g(x)}{h(x)}$ you get:

$$L=\lim_{x\to 0}f=\lim_{x\to 0}\frac{g'(x)}{h'(x)}$$ wich gives: $$L=lim_{x\to 0}\frac{\cos(x)-\cos(x)^2+\sin(x)^2}{2x}$$ If you apply l'Hopital again you get: $$L=\lim_{x\to 0}\frac{\sin(x)(4\cos(x)-1)}{2}=0$$

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  • $\begingroup$ Yeah, that makes a lot of sense, however I think my professor wants us to do it without using derivatives. $\endgroup$ – Steve Sep 25 '13 at 14:47
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Hint: Use the fundamental trigonometric limit: $$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$ For more informations about $\lim_{x\to 0}\frac{\sin x}{x}$ see Manipula Math Applets for $\lim_{x\to 0}\frac{\sin x}{x}$. For $\lim_{x\to 0}\frac{1-\cos x}{x}$, \begin{align} \lim_{x\to 0}\frac{1-\cos x}{x}=& \lim_{x\to 0}\frac{1-\cos x}{x}\frac{1+\cos x}{1+\cos x}\\ =& \lim_{x\to 0}\frac{1-\cos ^2x}{x\cdot(1+\cos x)}\\ =& \lim_{x\to 0}\frac{\sin^2x}{x\cdot(1+\cos x)}\\ =& \lim_{x\to 0}\frac{\sin x}{x}\frac{\sin x}{(1+\cos x)}\\ =& \lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x \to 0}\frac{1}{(1+\cos x)}\cdot \lim_{x\to 0}(\sin x)\\ =& 1\cdot\frac{1}{2}\cdot0 \end{align}

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